What does the evidence in this passage suggest?

A builder drops a brick from a height of 15 m above the ground. The gravitational field strength g is 10 N/ kg. What is the spee

Basile [38]

The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

Initial Velocity; $u = 0$

Height from which it has dropped; $h = 15m$

Gravitational field strength; $g = 10N/kg = 10 \frac{kg.m/s^2}{kg} = 10m/s^2$

Final speed of brick as it hits the ground; $v = \ ?$

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

$v^2 = u^2 + 2gh$

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

$v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s$

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Learn more about equations of motion: brainly.com/question/18486505

8 0

2 years ago

A sphere of radius r = 5cm carries a uniform volume charge density rho = 400 nC/m^3. Q. What is the total charge Q of the sphere

Tanzania [10]

Answer:

The total charge Q of the sphere is $2.094\times10^{-10}\ C$ .

Explanation:

Given that,

Radius = 5 cm

Charge density $J= 400\ nC/m^3$

We need to calculate the total charge Q of the sphere

Using formula of charge

$q=\rho V$

Where, $\rho$ = charge density

V = volume

Put the value into the formula

$q=\rho\times(\dfrac{4}{3}\pi r^3)$

Put the value into the formula

$q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3$

$q=2.094\times10^{-10}\ C$

Hence, The total charge Q of the sphere is $2.094\times10^{-10}\ C$ .

6 0

3 years ago

What is the required force to bring a 1550 kg vehicle moving at 32 m/s to a stop in a distance of 45 m?

Vilka [71]

The answer is 9 i think.

8 0

3 years ago

The strength of the force of friction depends on which two factors?

Stolb23 [73]

Options A and D are correct. The strength of the force of friction depends on the objects' sizes and weights and the heat generated by the friction and the types of surfaces involved.

<h3 /><h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and acceleration in the different components and balancing the equation gets. Components in the x-direction.

The strength of the force of friction depends on the two factors, as;

A. The objects' sizes and weights.

D. The heat generated by the friction and the types of surfaces involved.

Hence, options A and D are correct.

To learn more about the friction force, refer to the link;

brainly.com/question/1714663

#SPJ1

3 0

2 years ago

Consider to cars traveling directly towards each other. Car A has twice the mass of Car B. In order for the cars to completely s

Greeley [361]

Please mark brainliest you answer is B).

Have a good night!

7 0

3 years ago